2 of these questions should be "easy" to answer. how can you use symmetry to reduce the computation involved in the last 2?

at some point you will have to subtract the area of a "quarter-circle" from one of the "quarter-squares". do you know how to compute the area of a circle, if you know its radius?

there are a lot of lines and curves in that picture. not all of them are relevant to the solution. what *is* relevant is that you are told some of the curves are "semi-circles" (1/2 a circle), and that they are all the same size.

Aug 12th 2012, 07:33 PM

bjhopper

Re: Circles

Finding shaded area in the lower left square is the difficult part of problem.Label square from top left to lower left clockwise ABCD, Intersection of circles centered at A and B meet at P. P is on the perpendicular bisector of AB and DC. DPC is not shaded but is equal to the shaded. Draw a line parallel to DC thru P meeting AD at R and BC at S
Circle centered at A (0,0) X^2 + y^2 = 36 X =3 y = 3 rad3 M is midpoint of AB and N midpoint of DC. MP = 3rad3 NP = 6-3rad3.
Area of DPC = 6*(6-3rad3) - segment of 6cm circle whose height is 6-3rad3, chord 6 sector angle 60 deg

Aug 12th 2012, 09:35 PM

Deveno

Re: Circles

Quote:

Originally Posted by bjhopper

Finding shaded area in the lower left square is the difficult part of problem.Label square from top left to lower left clockwise ABCD, Intersection of circles centered at A and B meet at P. P is on the perpendicular bisector of AB and DC. DPC is not shaded but is equal to the shaded. Draw a line parallel to DC thru P meeting AD at R and BC at S
Circle centered at A (0,0) X^2 + y^2 = 36 X =3 y = 3 rad3 M is midpoint of AB and N midpoint of DC. MP = 3rad3 NP = 6-3rad3.
Area of DPC = 6*(6-3rad3) - segment of 6cm circle whose height is 6-3rad3, chord 6 sector angle 60 deg